Number Properties and Arithmetic Practice Questions — GMAT — Page 8

Question 141Question →

What is the smallest positive integer $n$ such that $180 \times n$ is a perfect square and $450 \times n$ is a perfect cube?

Show answer & explanation

Answer: 1620

Answer

The smallest positive integer

n

is 1620.

By writing 180 as

2^2 \times 3^2 \times 5^1

and 450 as

2^1 \times 3^2 \times 5^2

, we can analyze the exponents needed for

n = 2^a \times 3^b \times 5^c

. To make

180n

a perfect square,

a

and

b

must be even and

c

must be odd. To make

450n

a perfect cube,

1+a

,

2+b

, and

2+c

must be multiples of 3. Finding the minimal non-negative integers that satisfy both conditions yields

a = 2

,

b = 4

, and

c = 1

. Thus,

n = 2^2 \times 3^4 \times 5^1 = 1620

.

Step-by-Step Solution

1

Express 180 and 450 in terms of their prime factorizations.

180 = 2^2 \times 3^2 \times 5^1

and

450 = 2^1 \times 3^2 \times 5^2

Prime factorization exposes the exponent requirements for perfect powers.

2

Determine the constraints on exponents of

n = 2^a \times 3^b \times 5^c

.

For

180n

to be a square,

a

must be even,

b

must be even, and

c

must be odd. For

450n

to be a cube,

1+a

,

2+b

, and

2+c

must be multiples of 3.

A number is a perfect square if all prime exponents are even, and a perfect cube if all prime exponents are multiples of 3.

3

Find the minimal values for

a, b, c

.

a = 2, b = 4, c = 1

a=2

is even and makes

1+2=3

;

b=4

is even and makes

2+4=6

;

c=1

is odd and makes

2+1=3

.

4

Compute

n

.

n = 2^2 \times 3^4 \times 5^1 = 1620

Multiplying the prime powers together yields the smallest integer

n

.

Key Concept

Prime Factorization and Exponent Rules for Perfect Powers

Question 142Question →

An integer $N$ has exactly three distinct prime factors $p$ , $q$ , and $r$ , such that $N = p^a \cdot q^b \cdot r^c$ , where $a$ , $b$ , and $c$ are positive integers with $a < b < c$ . If $N$ has $24$ positive factors and $N^2$ has $105$ positive factors, what is the value of $a + b + c$ ?

Show answer & explanation

Answer: 6

Answer

The value of

a + b + c

is

6

.

For an integer with prime factorization

p^a q^b r^c

, the number of positive divisors is

(a+1)(b+1)(c+1)

. For

N^2 = p^{2a} q^{2b} r^{2c}

, the number of positive divisors is

(2a+1)(2b+1)(2c+1)

. Factoring

105

into three odd terms greater than

1

yields

3 \times 5 \times 7

. Matching these terms in increasing order gives

2a+1=3 \implies a=1

,

2b+1=5 \implies b=2

, and

2c+1=7 \implies c=3

. Checking

(1+1)(2+1)(3+1) = 24

confirms these values. Summing

a+b+c

yields

1+2+3=6

.

Step-by-Step Solution

1

Write the formula for the number of positive factors of

N

and

N^2

The number of positive factors of

N = p^a q^b r^c

is

(a+1)(b+1)(c+1) = 24

. Since

N^2 = p^{2a} q^{2b} r^{2c}

, the number of positive factors of

N^2

is

(2a+1)(2b+1)(2c+1) = 105

.

The number of divisors of a prime-factored integer is found by adding 1 to each exponent in its prime factorization and multiplying the results.

2

Factor

105

into three odd factors greater than

1

The prime factorization of

105

is

3 \times 5 \times 7

. The only way to express

105

as a product of three integers greater than

1

is

3 \times 5 \times 7

.

Since

a, b, c

are positive integers, each term

2a+1, 2b+1, 2c+1

must be an odd integer greater than

1

.

3

Assign the factors using the inequality condition

a < b < c

Since

a < b < c

, it follows that

2a+1 < 2b+1 < 2c+1

. Therefore:

2a+1 = 3 \implies a = 1

;

2b+1 = 5 \implies b = 2

;

2c+1 = 7 \implies c = 3

.

Matching the ordered values of

2a+1, 2b+1, 2c+1

with the sorted factors

3, 5, 7

uniquely determines

a, b,

and

c

.

4

Verify with the factor count for

N

and calculate the sum

(1+1)(2+1)(3+1) = 2 \times 3 \times 4 = 24

, which matches the given condition. The sum

a + b + c = 1 + 2 + 3 = 6

.

Verification confirms the solution satisfies all constraints.

Key Concept

Prime Factorization and Divisor Counting Formula

Estimated Time:1m 30s

Question 143Question →

If $a$ , $b$ , and $c$ are integers such that $a^2b + b^2c + c^2a$ is an odd integer, then the product $(a+b)(b+c)(c+a)$ must be an even integer.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the sum

(a+b) + (b+c) + (c+a) = 2(a+b+c)

is always an even integer, which mathematically prevents all three factors from being odd simultaneously.

The statement is true. The sum of the three factors

(a+b) + (b+c) + (c+a) = 2(a+b+c)

is guaranteed to be an even integer. If the product

(a+b)(b+c)(c+a)

were odd, each individual factor would have to be odd. But the sum of three odd integers is always odd, contradicting the fact that

2(a+b+c)

is even.

Step-by-Step Solution

1

Analyze the parity of

a^2b + b^2c + c^2a

using parity rules for exponents.

Since

x^2

has the same parity as

x

for any integer

x

,

a^2b + b^2c + c^2a

has the same parity as

ab + bc + ca

. Thus,

ab + bc + ca

is an odd integer.

Squaring an integer preserves its parity.

2

Determine possible parity combinations for

a

,

b

, and

c

.

For

ab + bc + ca

to be odd, either exactly two of

\{a, b, c\}

are odd, or all three are odd.

If 0 or 1 variable is odd, every product term (

ab, bc, ca

) is even, making the total sum even.

3

Evaluate the parity of the factors

(a+b)

,

(b+c)

, and

(c+a)

.

If two variables are odd and one is even, the sum of the two odd variables produces an even factor. If all three variables are odd, the sum of any two of them produces an even factor (all three factors become even).

The sum of two odd integers is always even.

4

Determine the parity of the product

(a+b)(b+c)(c+a)

.

Because at least one factor in the product is even under all valid scenarios, the product

(a+b)(b+c)(c+a)

must be an even integer.

Any product of integers containing at least one even factor is even.

Key Concept

Parity preservation under exponents and algebraic parity constraints of sums and products of integers.

Question 144Question →

If $x$ , $y$ , and $z$ are real numbers such that $x < 0 < y < z$ and $|x| > z$ , which of the following expressions MUST be positive?

\frac{x y}{x + z}

(x + z)(y - x)

\frac{x + y}{z}

(z - y)(x + z)

x + y + z

Show answer & explanation

Answer: $\frac{x y}{x + z}$

Answer

\frac{x y}{x + z}

Since

x < 0

and

y > 0

, the numerator

x y

is negative. From

x < 0

and

|x| > z

, we know

-x > z

, which means

x + z < 0

. Dividing the negative numerator by the negative denominator results in a strictly positive quotient. Thus, the expression stating 'x y over (x + z)' MUST be positive.

Step-by-Step Solution

1

Determine the sign of the numerator

x y

Since

x

is negative (

x < 0

) and

y

is positive (

y > 0

), their product

x y < 0

(negative).

The product of numbers with opposite signs is negative.

2

Determine the sign of the expression

x + z

using the absolute value condition

Since

x < 0

,

|x| = -x

. The inequality

|x| > z

becomes

-x > z

, which rearranges to

x + z < 0

(negative).

Subtracting

x

from both sides of

-x > z

yields

0 > x + z

.

3

Evaluate the quotient

\frac{x y}{x + z}

\frac{\text{negative}}{\text{negative}} = \text{positive}

.

Dividing two negative numbers yields a positive value.

Key Concept

Sign determination of products, quotients, and sums involving absolute values and inequalities

Question 145Question →

What is the smallest positive integer greater than $100$ that is divisible by $9$ ?

Show answer & explanation

Answer: 108

Answer

The smallest positive integer greater than

100

that is divisible by

9

is

108

.

An integer is divisible by 9 if it is an integer multiple of 9. Multiples of 9 around 100 are

9 \times 11 = 99

and

9 \times 12 = 108

. Since 99 is not greater than 100, the smallest integer meeting the condition is 108.

Step-by-Step Solution

1

Divide 100 by 9 to locate the adjacent multiples of 9.

100 = 9 \times 11 + 1

This shows that

9 \times 11 = 99

is the largest multiple of 9 that is less than or equal to 100.

2

Determine the next multiple of 9 by adding 9 to 99 (or evaluating

9 \times 12

).

99 + 9 = 108

Since 99 is less than 100, the next consecutive multiple of 9 must be the smallest multiple strictly greater than 100.

Key Concept

Divisibility and Multiples of Integers

Question 146Question →

If $p$ , $q$ , and $r$ are integers such that $3p + 2q + r^2 = 17$ , and $p$ is an odd integer, which of the following statements MUST be true?

r

is an even integer

q

must be an odd integer

p + r

is an even integer

r^2 + 1

is an even integer

p \cdot q \cdot r

must be an odd integer

Show answer & explanation

Answer: $r$ is an even integer

Answer

r

is an even integer

Because

p

is odd,

3p

is odd. The term

2q

is always even for any integer

q

. Adding an odd and an even number gives an odd sum (

3p + 2q

). Since the total sum

3p + 2q + r^2

equals

17

, which is an odd number,

r^2

must be even. An integer whose square is even must be even itself, so the statement that

r

is an even integer must be true.

Step-by-Step Solution

1

Analyze the parity of the term

3p

Since

p

is given as an odd integer,

\text{odd} \times \text{odd} = \text{odd}

, so

3p

is an odd integer.

Multiplying an odd integer by an odd integer yields an odd integer.

2

Analyze the parity of the term

2q

2q

is an even integer for any integer

q

.

Any integer multiplied by

2

produces an even integer.

3

Determine the parity of the sum

3p + 2q

3p + 2q = \text{odd} + \text{even} = \text{odd}

.

The sum of an odd integer and an even integer is always odd.

4

Evaluate the equation

(3p + 2q) + r^2 = 17

to find the parity of

r^2

and

r

\text{odd} + r^2 = 17\ (\text{odd}) \implies r^2

must be even

\implies r

must be an even integer.

For the sum of an odd integer and

r^2

to equal an odd integer (

17

),

r^2

must be even. An integer whose square is even must itself be even.

Key Concept

Parity rules of addition and multiplication of integers:

\text{odd} + \text{even} = \text{odd}

and

\text{even} \times \text{any integer} = \text{even}

.

Estimated Time:1m 30s

Question 147Question →

The original price of a sweater is $\$40$ . If the price is increased by $25\%$ , and then the new price is decreased by $10\%$ , what is the final price of the sweater?

\$44

\$45

\$46

\$50

\$36

Show answer & explanation

Answer: $\$45$

Answer

The final price of the sweater is

\$45

.

Increasing the original price of

\$40

by

25\%

yields an intermediate price of

\$50

(since

25\%

of

\$40

is

\$10

). Decreasing

\$50

by

10\%

reduces the price by

\$5

(since

10\%

of

\$50

is

\$5

), resulting in a final price of

\$45

.

Step-by-Step Solution

1

Calculate the price after the initial

25\%

increase.

The price increases by

25\%

of

\$40

, which is

0.25 \times \$40 = \$10

. The new intermediate price is

\$40 + \$10 = \$50

.

Percentage increases must be calculated using the starting base value.

2

Calculate the final price after the

10\%

decrease.

The discount is

10\%

of the intermediate price of

\$50

, which is

0.10 \times \$50 = \$5

. Subtracting this from the intermediate price gives

\$50 - \$5 = \$45

.

Successive percentage changes apply to the updated base value, not the original starting value.

Key Concept

Successive Percentage Changes and Base Value Alignment

Question 148Question →

What is the smallest positive integer $n$ such that $n$ is divisible by both $12$ and $18$ , and $n$ has exactly $12$ positive divisors?

Show answer & explanation

Answer: 72

Answer

The smallest positive integer satisfying all given conditions is 72.

To satisfy the divisibility requirements,

n

must be a multiple of

\text{LCM}(12, 18) = 36 = 2^2 \times 3^2

. Thus,

n = 2^a \times 3^b \times \dots

where

a \geq 2

and

b \geq 2

. The total number of positive divisors of

n

is given by

(a+1)(b+1)\dots = 12

. Since

a+1 \geq 3

and

b+1 \geq 3

,

n

cannot have any additional prime factors, as that would yield at least

3 \times 3 \times 2 = 18

divisors. Factoring 12 into two integers each at least 3 gives

3 \times 4

or

4 \times 3

. Setting

(a+1, b+1) = (3, 4)

yields

a = 2, b = 3

, giving

n = 2^2 \times 3^3 = 108

. Setting

(a+1, b+1) = (4, 3)

yields

a = 3, b = 2

, giving

n = 2^3 \times 3^2 = 72

. The smallest value is 72.

Step-by-Step Solution

1

Determine the prime factor constraints on

n

n

must be divisible by

2^2

and

3^2

, so

n = 2^a \times 3^b \times \dots

with

a \geq 2

and

b \geq 2

.

For

n

to be divisible by

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

, its prime factorization must contain at least the maximum power of each prime appearing in either 12 or 18.

2

Set up the divisor count equation

(a+1)(b+1) = 12

with

a+1 \geq 3

and

b+1 \geq 3

, and no additional prime factors.

The number of divisors of

n = p_1^{e_1} p_2^{e_2} \dots

is

(e_1+1)(e_2+1)\dots

. Since

(a+1) \geq 3

and

(b+1) \geq 3

, adding another prime factor would make the product at least

3 \times 3 \times 2 = 18 > 12

.

3

Find all possible pairs of exponents

(a, b)

and calculate candidate values for

n

Candidate 1:

a=2, b=3 \implies n = 2^2 \times 3^3 = 108

. Candidate 2:

a=3, b=2 \implies n = 2^3 \times 3^2 = 72

.

The only factor pairs of 12 into integers

\geq 3

are

3 \times 4

and

4 \times 3

.

4

Identify the minimum value

n = 72

72 < 108

, so 72 is the smallest positive integer fulfilling all criteria.

Key Concept

Divisor Count Formula & Least Common Multiple constraints

Question 149Question →

For any integer $n$ , the expression $n^2 + n + 7$ is an odd integer.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because

n(n + 1)

is the product of two consecutive integers, which is always even. Adding the odd constant

7

to an even integer yields an odd integer for all integer values of

n

.

The statement is true because

n^2 + n = n(n + 1)

represents the product of two consecutive integers and is therefore always even. Adding

7

(an odd integer) to an even integer always yields an odd integer regardless of whether

n

is positive, negative, or zero.

Step-by-Step Solution

1

Factor the algebraic terms

n^2 + n

.

n^2 + n + 7 = n(n + 1) + 7

Factoring out

n

reveals the product of two consecutive integers.

2

Determine the parity of the product of two consecutive integers

n(n + 1)

.

n(n + 1)

is always even.

Of any two consecutive integers, exactly one is even. The product of an even integer and any integer is always even.

3

Determine the overall parity of the expression by adding

7

.

\text{even} + 7 = \text{odd}

Adding an odd integer to an even integer results in an odd integer.

Key Concept

The product of two consecutive integers is always even. Adding an odd integer to an even integer results in an odd integer.

Estimated Time:45s

Question 150Question →

What is the units digit of $2^{35}$ ?

2

4

6

8

0

Show answer & explanation

Answer: 8

Answer

The units digit of

2^{35}

is 8.

The units digit of powers of 2 repeats every 4 powers in the sequence 2, 4, 8, 6. Dividing the exponent 35 by 4 yields a quotient of 8 and a remainder of 3. A remainder of 3 corresponds to the third number in the repeating sequence, which is 8.

Step-by-Step Solution

1

Find the cyclicity pattern of the units digits for powers of 2.

The units digits of

2^1, 2^2, 2^3, 2^4, 2^5, \dots

are

2, 4, 8, 6, 2, \dots

, repeating in a cycle of length 4.

Units digits of powers follow a repeating periodic pattern determined by base arithmetic modulo 10.

2

Divide the exponent by the length of the cycle.

35 \div 4 = 8

with a remainder of

3

.

The remainder indicates how far into the 4-term repeating cycle the exponent 35 reaches.

3

Match the remainder to the corresponding term in the units digit cycle.

A remainder of 3 corresponds to the 3rd term in the cycle

(2, 4, 8, 6)

, which is 8.

The 3rd power in the repeating pattern gives a units digit of

2^3 = 8

.

Key Concept

Units Digit Cyclicity

Estimated Time:1m 0s

Question 151Question →

If $a$ , $b$ , and $c$ are integers such that $a^2 + 3b$ is an odd integer, $b^2 + 5c$ is an even integer, and $c^2 + 7a$ is an odd integer, which of the following expressions MUST be an even integer?

a(b + c)

a + b + c

a + bc

ab + c

ac + b

Show answer & explanation

Answer: $a(b + c)$

Answer

The expression

a(b + c)

MUST be an even integer.

The expression

a(b + c)

is guaranteed to be an even integer because in both valid parity configurations—whether

a

is odd with

b, c

even, or

a

is even with

b, c

odd—the sum

(b + c)

is always even, making the product

a(b + c)

even.

Step-by-Step Solution

1

Analyze the parity relationships from the given expressions.

Since

a^2

has the same parity as

a

,

3b

has the same parity as

b

,

b^2

has the same parity as

b

,

5c

has the same parity as

c

,

c^2

has the same parity as

c

, and

7a

has the same parity as

a

, the given conditions simplify to: (1)

a + b

is odd, (2)

b + c

is even, and (3)

c + a

is odd.

Multiplying an integer by an odd constant or raising an integer to a positive integer power does not change its parity.

2

Determine the valid parity configurations for

a

,

b

, and

c

.

Condition (1) implies

a

and

b

have opposite parities. Condition (2) implies

b

and

c

have the same parity. Condition (3) implies

c

and

a

have opposite parities. This yields two valid cases: Case 1:

b

and

c

are even, while

a

is odd. Case 2:

b

and

c

are odd, while

a

is even.

A sum of two integers is odd when they have opposite parities, and even when they have the same parity.

3

Evaluate the target expression

a(b + c)

across both valid cases.

In Case 1 (

a

odd,

b

even,

c

even):

b + c

is even, so

a(b + c) = \text{odd} \times \text{even} = \text{even}

. In Case 2 (

a

even,

b

odd,

c

odd):

b + c

is even, so

a(b + c) = \text{even} \times \text{even} = \text{even}

.

Any integer multiplied by an even integer yields an even integer.

Key Concept

Systematic parity deduction across multiple constraints and testing expressions under all valid parity configurations.

Question 152Question →

What is the smallest positive integer greater than $1$ that leaves a remainder of $1$ when divided by each of $4$ , $6$ , and $8$ ?

Show answer & explanation

Answer: 25

Answer

The correct answer is 25.

The least common multiple of 4, 6, and 8 is 24, which is the smallest positive integer divisible by all three numbers. To leave a remainder of 1 upon division by each, we add 1 to 24, resulting in 25.

Step-by-Step Solution

1

Calculate the least common multiple (LCM) of 4, 6, and 8.

\text{LCM}(4, 6, 8) = 24

Any positive integer divisible by 4, 6, and 8 must be a multiple of their least common multiple.

2

Add the desired remainder of 1 to the LCM.

24 + 1 = 25

Adding 1 to a common multiple guarantees that division by 4, 6, or 8 yields a remainder of 1.

Key Concept

Least Common Multiple (LCM) and Remainder Properties

Question 153Question →

Two positive integers $a$ and $b$ are in the ratio $3 : 8$ . If the least common multiple (LCM) of $a$ and $b$ is $360$ , what is the greatest common divisor (GCD) of $a$ and $b$ ?

Show answer & explanation

Answer: 15

Answer

The greatest common divisor of

a

and

b

is 15.

Any two positive integers in the ratio

3 : 8

can be written as

3g

and

8g

, where

g

is their greatest common divisor. Because

3

and

8

are coprime (their GCD is

1

), the least common multiple of

3g

and

8g

is

3 \times 8 \times g = 24g

. Setting

24g = 360

and dividing by

24

yields

g = 15

.

Step-by-Step Solution

1

Represent the two integers using their ratio and their greatest common divisor.

Let

g = \gcd(a, b)

. Then

a = 3g

and

b = 8g

, where

3

and

8

share no common factors other than

1

.

When two numbers are in reduced ratio

p : q

, dividing both by their GCD leaves coprime factors

p

and

q

.

2

Express the LCM of

a

and

b

in terms of

g

.

\text{LCM}(a, b) = 3 \times 8 \times g = 24g.

The LCM of two numbers

p \cdot g

and

q \cdot g

with

\gcd(p, q) = 1

is

p \cdot q \cdot g

.

3

Solve for

g

using the given LCM value of

360

.

24g = 360 \implies g = 15.

Dividing the given LCM by the product of the coprime ratio components yields the GCD.

Key Concept

Relationship between GCD, LCM, and coprime factor ratios of two positive integers

Estimated Time:1m 15s

Question 154Question →

A positive integer $n$ has only two distinct prime factors, $2$ and $3$ . If $n$ is a multiple of $12$ , is not divisible by $8$ , and has exactly $12$ positive divisors, what is the value of $n$ ?

Show answer & explanation

Answer: 108

Answer

The value of

n

is 108.

Because

n

has only

2

and

3

as prime factors, its prime factorization is

n = 2^a \times 3^b

. Divisibility by

12 = 2^2 \times 3^1

requires

a \ge 2

, and non-divisibility by

8 = 2^3

requires

a < 3

, which forces

a = 2

. Using the formula for total positive divisors,

(2+1)(b+1) = 12

, which simplifies to

3(b+1) = 12

and gives

b = 3

. Calculating

n = 2^2 \times 3^3

yields

108

.

Step-by-Step Solution

1

Set up the prime factorization of

n

n = 2^a \times 3^b

where

a \ge 1

and

b \ge 1

The problem states that

2

and

3

are the only distinct prime factors of

n

.

2

Determine the exact value of exponent

a

a = 2

n

is divisible by

12 = 2^2 \times 3^1

(so

a \ge 2

) but not by

8 = 2^3

(so

a < 3

).

3

Determine the exact value of exponent

b

b = 3

The total number of positive divisors is

(a+1)(b+1) = (2+1)(b+1) = 3(b+1) = 12

, which solves to

b = 3

.

4

Calculate the value of

n

n = 108

n = 2^2 \times 3^3 = 4 \times 27 = 108

.

Key Concept

Determining integer values using prime factorization and the number of divisors formula

Question 155Question →

What is the sum of the distinct prime factors of $60$ ?

Show answer & explanation

Answer: 10

Answer

The sum of the distinct prime factors of 60 is 10.

The prime factorization of

60

is

2^2 \times 3 \times 5

. The unique prime factors are

2

,

3

, and

5

. Adding these values together gives

2 + 3 + 5 = 10

.

Step-by-Step Solution

1

Find the prime factorization of 60

60 = 2^2 \times 3^1 \times 5^1

Decompose 60 into prime factor powers.

2

List the distinct prime factors

The distinct prime factors are 2, 3, and 5

Focus only on the prime bases, ignoring exponents.

3

Sum the distinct prime factors

2 + 3 + 5 = 10

Add the unique prime numbers found in the factorization.

Key Concept

Prime Factorization and Distinct Prime Factors

Question 156Question →

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $14$ and a least common multiple (LCM) of $840$ . If $14 < x < y < 200$ and $x$ is not a multiple of $4$ , what is the value of $y - x$ ?

42

98

112

126

238

Show answer & explanation

Answer: $98$

Answer

The value of

y - x

is

98

.

The prime factorizations of

\text{GCD}(x,y)=14=2^1 \times 7^1

and

\text{LCM}(x,y)=840=2^3 \times 3^1 \times 5^1 \times 7^1

require

x

and

y

to split the exponents of

2, 3, 5, 7

. Since

x

is not a multiple of

4

,

x

contains

2^1

and

y

contains

2^3

. Both contain

7^1

. To satisfy

14 < x < y < 200

,

x

must take factor

5

(

x = 70

) and

y

must take factor

3

(

y = 168

). The difference is

168 - 70 = 98

.

Step-by-Step Solution

1

Express the GCD and LCM in prime factorized form.

\text{GCD}(x, y) = 14 = 2^1 \times 3^0 \times 5^0 \times 7^1

and

\text{LCM}(x, y) = 840 = 2^3 \times 3^1 \times 5^1 \times 7^1

.

Prime factorization allows us to analyze the min and max prime exponents shared between

x

and

y

.

2

Determine the distribution of prime factor exponents between

x

and

y

.

For factor

7

: both

x

and

y

must have exponent

1

.
For factor

2

:

\min(a_x, a_y) = 1

and

\max(a_x, a_y) = 3

. Since

x

is not divisible by

4

(

2^2

),

a_x = 1

, which forces

a_y = 3

.
For factors

3

and

5

: one integer receives

3^1

and the other receives

5^1

.

The GCD takes the minimum exponent and the LCM takes the maximum exponent for each prime factor.

3

Test possible assignments of factors

3

and

5

under the given inequalities.

Case 1:

x

gets

3

and

y

gets

5

.

x = 2 \times 7 \times 3 = 42

,

y = 8 \times 7 \times 5 = 280

. This fails

y < 200

.
Case 2:

x

gets

5

and

y

gets

3

.

x = 2 \times 7 \times 5 = 70

,

y = 8 \times 7 \times 3 = 168

. This satisfies

14 < 70 < 168 < 200

.

Only Case 2 satisfies all structural and boundary constraints (

14 < x < y < 200

and

x

not a multiple of

4

).

4

Compute

y - x

.

y - x = 168 - 70 = 98

.

Direct subtraction of the valid integer values.

Key Concept

Prime Factorization Rules for GCD and LCM

Question 157Question →

Let $n$ be a positive integer whose only prime factors are $2$ , $3$ , and $5$ . If $\dfrac{n}{2}$ has $36$ positive divisors, $\dfrac{n}{3}$ has $36$ positive divisors, and $\dfrac{n}{5}$ has $32$ positive divisors, how many positive divisors does $n^2$ have?

96

180

216

245

343

Show answer & explanation

Answer: 245

Answer

245

Writing

n

as

2^a \cdot 3^b \cdot 5^c

, the number of divisors of

\frac{n}{2}

,

\frac{n}{3}

, and

\frac{n}{5}

leads to the system

a(b+1)(c+1) = 36

,

(a+1)b(c+1) = 36

, and

(a+1)(b+1)c = 32

. Equating the first two yields

a = b

. Substituting

b = a

into the third gives

(a+1)^2 c = 32

. Since

a

is a positive integer,

(a+1)^2

must be a square dividing

32

, so

a+1 = 4 \implies a = 3

, giving

b = 3

and

c = 2

. Therefore,

n^2 = 2^6 \cdot 3^6 \cdot 5^4

, and the total number of positive divisors of

n^2

is

(6+1)(6+1)(4+1) = 245

.

Step-by-Step Solution

1

Express

n

in terms of its prime factorization

n = 2^a \cdot 3^b \cdot 5^c

where

a, b, c \ge 1

The problem states that

2

,

3

, and

5

are the only prime factors of

n

.

2

Set up equations for the number of divisors of

\dfrac{n}{2}

,

\dfrac{n}{3}

, and

\dfrac{n}{5}

a(b+1)(c+1) = 36

,

(a+1)b(c+1) = 36

, and

(a+1)(b+1)c = 32

Dividing

n

by

2

,

3

, or

5

decreases the corresponding prime exponent by

1

. The formula for total positive divisors of

2^p 3^q 5^r

is

(p+1)(q+1)(r+1)

.

3

Solve the system of equations for exponents

a

,

b

, and

c

a = 3

,

b = 3

,

c = 2

Since

a(b+1)(c+1) = (a+1)b(c+1) = 36

, we get

a(b+1) = b(a+1) \implies a = b

. Substituting

b = a

into

(a+1)^2 c = 32

gives

(a+1)^2

as a factor of

32

. Testing perfect squares yields

a+1 = 4 \implies a = 3

, so

c = 2

. Checking in the first equation gives

3(4)(3) = 36

, which is consistent.

4

Determine the prime factorization and number of positive divisors of

n^2

n^2 = 2^6 \cdot 3^6 \cdot 5^4

, total divisors

= (6+1)(6+1)(4+1) = 7 \cdot 7 \cdot 5 = 245

Squaring

n

doubles each exponent in its prime factorization.

Key Concept

Prime Factorization and Number of Divisors Formula

Estimated Time:2m 0s

Question 158Question →

For a positive integer $N$ , the greatest common divisor of $N$ and $360$ is $120$ , and the least common multiple of $N$ and $450$ is $9000$ . What is the value of $N$ ?

Show answer & explanation

Answer: 3000

Answer

The value of

N

is

3000

.

Prime factorizing

360

,

120

,

450

, and

9000

converts the GCD and LCM requirements into a system of min/max equations for the exponents of

2

,

3

, and

5

. The condition

\gcd(N, 360) = 120

dictates that the exponent of

3

in

N

must be exactly

1

, while the exponent of

2

is at least

3

. The condition

\text{lcm}(N, 450) = 9000

dictates that the exponent of

2

must be exactly

3

and the exponent of

5

must be exactly

3

. Combining these constraints gives

N = 2^3 \times 3^1 \times 5^3 = 3000

.

Step-by-Step Solution

1

Find the prime factorizations of all known numbers

360 = 2^3 \times 3^2 \times 5^1

,

120 = 2^3 \times 3^1 \times 5^1

,

450 = 2^1 \times 3^2 \times 5^2

, and

9000 = 2^3 \times 3^2 \times 5^3

.

Prime factorization allows us to analyze GCD and LCM conditions using exponent minimums and maximums.

2

Apply the GCD exponent rule

\min(\text{exp}_N(p), \text{exp}_A(p)) = \text{exp}_{GCD}(p)

For factor

2

:

\min(a, 3) = 3 \implies a \ge 3

; For factor

3

:

\min(b, 2) = 1 \implies b = 1

; For factor

5

:

\min(c, 1) = 1 \implies c \ge 1

.

The GCD of two numbers takes the minimum exponent for each prime factor.

3

Apply the LCM exponent rule

\max(\text{exp}_N(p), \text{exp}_B(p)) = \text{exp}_{LCM}(p)

For factor

2

:

\max(a, 1) = 3 \implies a = 3

; For factor

5

:

\max(c, 2) = 3 \implies c = 3

. No prime factors greater than

5

exist in

N

.

The LCM of two numbers takes the maximum exponent for each prime factor.

4

Synthesize the exponents and compute

N

N = 2^3 \times 3^1 \times 5^3 = 8 \times 3 \times 125 = 3000

.

Multiplying out the uniquely determined prime factors gives the value of

N

.

Key Concept

Simultaneous prime exponent analysis using GCD (minimum exponents) and LCM (maximum exponents) rules.

Question 159Question →

A retail warehouse received a large shipment of electronics. On Monday, the warehouse sold $\frac{2}{5}$ of the total shipment. On Tuesday, it sold $25\%$ of the items that remained after Monday. On Wednesday, it sold $33\frac{1}{3}\%$ of the items that remained after Tuesday. If $120$ items remained unsold at the end of Wednesday, how many items were in the original shipment?

400

600

1,200

7,200

36

Show answer & explanation

Answer: $400$

Answer

The original shipment contained

400

items.

The correct option correctly evaluates the fraction remaining after each successive sale. After selling

\frac{2}{5}

on Monday,

\frac{3}{5}

remained. After selling

25\%

(

\frac{1}{4}

) on Tuesday,

\frac{3}{4}

of the previous remainder (

\frac{9}{20}

) remained. After selling

33\frac{1}{3}\%

(

\frac{1}{3}

) on Wednesday,

\frac{2}{3}

of that remainder remained, giving a net remaining fraction of

\frac{3}{10}

. Equating

\frac{3}{10}

of the total to

120

yields an original shipment size of

400

.

Step-by-Step Solution

1

Calculate the fraction of items remaining after Monday's sales.

Since

\frac{2}{5}

were sold,

1 - \frac{2}{5} = \frac{3}{5}

of the original shipment remained.

The remaining fraction is

1

minus the fraction sold.

2

Calculate the fraction of items remaining after Tuesday's sales.

Tuesday sold

25\% = \frac{1}{4}

of the remaining items, leaving

1 - \frac{1}{4} = \frac{3}{4}

of Monday's remainder. The fraction remaining relative to the original total is

\frac{3}{5} \times \frac{3}{4} = \frac{9}{20}

.

Successive percentage reductions apply to the updated intermediate remainder.

3

Calculate the fraction of items remaining after Wednesday's sales.

Wednesday sold

33\frac{1}{3}\% = \frac{1}{3}

of Tuesday's remainder, leaving

1 - \frac{1}{3} = \frac{2}{3}

of that remainder. The final fraction remaining relative to the original total is

\frac{9}{20} \times \frac{2}{3} = \frac{6}{20} = \frac{3}{10}

.

Multiply by the fraction remaining after Wednesday's reduction.

4

Set up the equation to solve for the original total number of items

N

.

\frac{3}{10}N = 120 \implies N = 120 \times \frac{10}{3} = 400

.

Divide the remaining count by the net remaining fraction.

Key Concept

Successive Percent and Fraction Reductions

Estimated Time:1m 30s

Question 160Question →

For two positive integers $x$ and $y$ with $x < y$ , the greatest common divisor is $\text{GCD}(x, y) = 15$ and the least common multiple is $\text{LCM}(x, y) = 9000$ . If $x$ is a multiple of $8$ but not a multiple of $9$ , what is the value of $y - x$ ?

Show answer & explanation

Answer: 1005

Answer

1005

By writing

x = 15a

and

y = 15b

with

\text{GCD}(a, b) = 1

, we derive

a \cdot b = \frac{9000}{15} = 600

. Prime factorizing

600 = 8 \times 3 \times 25

yields four coprime pairs

(a, b)

with

a < b

:

(1, 600)

,

(3, 200)

,

(8, 75)

, and

(24, 25)

. These yield candidate values for

x

of

15

,

45

,

120

, and

360

, respectively. Checking the divisibility conditions,

15

and

45

are not multiples of

8

, while

360

is a multiple of

9

. The only value of

x

that is a multiple of

8

and not a multiple of

9

is

120

(corresponding to

a = 8, b = 75

). Hence

y = 15 \times 75 = 1125

, and

y - x = 1125 - 120 = 1005

.

Step-by-Step Solution

1

Set up the algebraic representation using the GCD

x = 15a

and

y = 15b

, where

\text{GCD}(a, b) = 1

and

a < b

Any two integers can be written as the product of their GCD and coprime factor multipliers.

2

Relate LCM and GCD to find the product of multipliers

a \times b

a \times b = \frac{\text{LCM}(x, y)}{\text{GCD}(x, y)} = \frac{9000}{15} = 600

The product of GCD and LCM equals the product of the numbers:

\text{GCD}(x,y) \times \text{LCM}(x,y) = x \times y = 15a \times 15b = 225ab

.

3

Decompose 600 into coprime component blocks

600 = 2^3 \times 3^1 \times 5^2 = 8 \times 3 \times 25

Since

\text{GCD}(a, b) = 1

, prime powers cannot be split between

a

and

b

.

4

Form all valid candidate pairs

(a, b)

with

a < b

(1, 600)

,

(3, 200)

,

(8, 75)

, and

(24, 25)

There are

2^{3-1} = 4

ways to partition the 3 prime factor blocks into two coprime factors where

a < b

.

5

Apply divisibility constraints to isolate

x

and

y

x = 120

and

y = 1125

Only

x = 15 \times 8 = 120

satisfies being a multiple of 8 without being a multiple of 9.

6

Calculate the target difference

y - x

1125 - 120 = 1005

Subtracting

x

from

y

yields the required value.

Key Concept

Partitioning prime factor powers of LCM/GCD to identify coprime factor multipliers

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